JEE Main 2023 — Straight Lines Question with Solution

Plotting the diagram of the given data we get,

Equation of straight line

Here given $OA=a \& OB=b$, so equation of line $AB$ in intercept form will be $\frac{x}{a}+\frac{y}{b}=1 ......\left(1\right)$

And given perpendicular line of length $p$ is making $\frac{\mathrm{\pi }}{3}$ angle with the $x-$axis,

So, the equation of line in normal form will be

$x\cos \left(\frac{\mathrm{\pi }}{3}\right)+y\sin \left(\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 3}}\right)=p$

$\Rightarrow \frac{x}{2p}+\frac{y\sqrt{3}}{2p}=1$

So, on comparing with the equation $\left(1\right)$ we get,

$a=2p \& b=\frac{2p}{\sqrt{3}}$

Now area of the triangle $OAB$ is 

$\frac{1}{2}\times a\times b=\frac{98\sqrt{3}}{3}$

$\Rightarrow \frac{1}{2}\times 2p\times \frac{2p}{\sqrt{3}}=\frac{98\sqrt{3}}{3}$

$\Rightarrow p^2=49$

Therefore,

$a^2-b^2=4p^2-\frac{4p^2}{3}=\frac{8}{3}{p}^2$

$\Rightarrow a^2-b^2=\frac{8}{3}\left(49\right)=\frac{392}{3}$